Why Two Market Loss Estimates Can Disagree
A market VaR and a sensitivity-based loss estimate can both be calculated correctly and still differ. They may describe different horizons or confidence levels, or one may capture nonlinear valuation effects, market factors, or tail behavior that the other does not. The first task is therefore not to decide which number is “right,” but to identify what each number measures.
This tutorial focuses on diagnosing those differences. It assumes the reader can interpret market-factor sensitivities and VaR estimates, including the methods used in the earlier material on sensitivity-based estimates and parametric, historical simulation, and Monte Carlo VaR. The comparison principles here are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
The sign convention matters: \(L\) is positive when the portfolio loses value. A VaR of $5 million at 99% means that, under the specified model, the loss is at or below $5 million with probability at least 99%. VaR does not describe how large losses may be beyond that quantile.
Put the Estimates on a Common Basis
Before interpreting a gap, write down the assumptions behind each number. The following comparison dimensions are especially useful.
| Dimension | Questions to ask |
|---|---|
| Horizon | Do both estimates cover the same number of days or other time period? If a horizon was scaled, what assumptions justified that scaling? |
| Confidence level | Are the estimates the same quantile, such as 95% or 99%? A higher confidence level generally produces a larger VaR. |
| Valuation response | Does the sensitivity estimate use a first-order approximation, while the VaR model revalues nonlinear positions? |
| Factor coverage | Are rates, equity, foreign exchange, spreads, and other material factors represented in both estimates? |
| Distribution | Do the models use the same factor distributions and dependence assumptions, especially in the tail? |
For example, a first-order sensitivity calculation may approximate portfolio value change by \( \Delta V\approx \mathbf{g}^{\mathsf T}\Delta\mathbf{z}\), where \(\mathbf{g}\) is the vector of sensitivities and \(\Delta\mathbf{z}\) is the vector of market-factor changes. Its corresponding loss is approximately \(L\approx-\mathbf{g}^{\mathsf T}\Delta\mathbf{z}\). If the factors are jointly normal with mean zero and covariance matrix \(\Sigma\), then this linearized loss is normal with variance \(\mathbf{g}^{\mathsf T}\Sigma\mathbf{g}\). Thus, under those assumptions,
where \(z_{\alpha}\) is the \(\alpha\)-quantile of a standard normal distribution. If the modeled loss has mean \(\mu_L\) rather than zero, the corresponding expression is \(\mu_L+z_{\alpha}\sigma_L\), where \(\sigma_L\) is its standard deviation. These formulas do not generally apply to nonlinear valuation, nonnormal factors, or an estimate with a different horizon unless the model assumptions justify the required adjustments.
Why the Linear Normal Formula Holds
The formula follows from the distribution of a linear combination of jointly normal variables. Under the assumptions above, set \(L=-\mathbf{g}^{\mathsf T}\Delta\mathbf{z}\). Its mean is zero, and its variance is
A normal variable with mean zero and standard deviation \(\sigma_L\) has \(\alpha\)-quantile \(z_{\alpha}\sigma_L\): standardizing gives \(P(L\leq z_{\alpha}\sigma_L)=P(L/\sigma_L\leq z_{\alpha})=\alpha\). This establishes the stated expression. The argument depends on both normality and the linear approximation; a similar-looking calculation is not automatically a valid VaR when either condition fails.
Worked Comparisons
Worked Example: Aligning Horizon and Confidence Level
A hypothetical portfolio has a one-day loss that is normal with mean $0 and standard deviation $1.50 million. Assume daily losses are independent and identically distributed, and the sensitivity model is linear. Compare its 10-day sensitivity-based VaR with a reported 10-day, 95% VaR of $8.40 million. Use \(z_{0.95}=1.6449\) and \(z_{0.99}=2.3263\). Report dollar amounts in millions to the nearest $0.001 million.
Method. Calculate the sensitivity estimates at both stated confidence levels, using square-root-of-time scaling only because the example assumes independent, identically distributed normal daily losses and a linear exposure.
For 10 days, the standard deviation is \(1.50\sqrt{10}=4.743416\) million. The 10-day, 95% estimate is
At 99%, the sensitivity estimate is
The reported $8.40 million is at 10 days and 95%, so compare it with the 10-day, 95% sensitivity estimate, not with the 99% estimate. The reported result exceeds the sensitivity estimate by \(8.400000-7.802446=0.597554\) million, or $0.598 million to the nearest $0.001 million.
Check and trap avoided. The 99% estimate is larger because it uses a more extreme quantile; it is not the like-for-like comparator. Also, the square-root-of-time assumption is not a general rule for all VaR models: serial dependence, changing volatility, or material nonlinear effects can invalidate it.
Worked Example: Separating First-Order and Nonlinear Effects
A hypothetical equity position has value change \(\Delta V=5{,}000{,}000x-10{,}000{,}000x^2\), where the one-period market return \(x\) is uniform on \([-0.02,0.02]\). Find the exact 99% loss VaR under this mapping and compare it with the first-order sensitivity estimate. Report dollars to the cent.
Method. First find the factor value corresponding to the 99th percentile of loss. Since \(L=-\Delta V\), the loss as a function of return is \(L(x)=-5{,}000{,}000x+10{,}000{,}000x^2\). Its derivative is \(-5{,}000{,}000+20{,}000{,}000x\), which is negative throughout \([-0.02,0.02]\). Therefore, loss decreases as \(x\) increases: the 99th percentile of loss corresponds to the 1st percentile of \(x\).
For a uniform variable on an interval of width \(0.04\), the 1st percentile is \(-0.02+0.01(0.04)=-0.0196\). Substitution into the full valuation mapping gives
The first-order estimate drops the squared term, giving \(5{,}000{,}000(0.0196)=\$98{,}000.00\). Here the nonlinear term increases the 99% loss estimate by $3,841.60.
Check and trap avoided. The factor percentile is \(-0.0196\), not \(-0.0192\): the 1st percentile is one percent of the interval width above its lower endpoint. The monotonicity check also matters; without it, one could select the wrong factor tail for the loss quantile.
Worked Example: Detecting an Omitted Market Factor
A hypothetical portfolio has independent, zero-mean, normal one-day loss components from interest rates and foreign exchange, with standard deviations $0.60 million and $0.80 million, respectively. A sensitivity report includes only the rate component. Find the 99% VaR with both components and quantify the difference from the rate-only estimate. Use \(z_{0.99}=2.3263\). Report amounts in millions to the nearest $0.001 million.
Method. Under independence, the variance of the total loss is the sum of the component variances. The total standard deviation is
The VaR with both factors is \(2.3263(1.00)=\$2.326\) million. The rate-only estimate is \(2.3263(0.60)=\$1.39578\) million, or $1.396 million. The full-factor VaR is therefore \(2.3263-1.39578=0.93052\) million, or $0.931 million higher.
Check and trap avoided. The difference is not evidence that either calculation is arithmetically wrong: the estimates cover different factor sets. The independence assumption is essential to adding variances this way. If the components were dependent, covariance would also enter the total variance.
Worked Example: Testing a Distributional Assumption
A hypothetical one-period loss \(L\), measured in millions of dollars, is uniform on \([-1,1]\). Compare its exact 99% VaR with a normal estimate that uses the same mean and variance. Use \(z_{0.99}=2.3263\). Report amounts to the nearest $0.001 million.
Method. The 99th percentile of a uniform distribution on \([-1,1]\) is \(-1+0.99(2)=0.98\), so the exact VaR is $0.980 million. The uniform mean is zero and its variance is \((2)^2/12=1/3\) million-squared. The standard deviation is \(1/\sqrt{3}=0.577350\) million. A moment-matched normal estimate is therefore
The normal estimate is $0.363 million above the exact uniform quantile. It even exceeds the maximum possible loss of $1 million under the stated uniform model. That is possible because the normal approximation assigns probability to losses beyond the uniform model’s support; it is not the exact VaR for the stated distribution.
Check and trap avoided. Matching the first two moments does not match the tail quantiles. The comparison is between two different distributional models, not two exact calculations of the same quantile.
A Practical Diagnostic Sequence
A disciplined review helps separate harmless presentation differences from material model differences. Apply the checks in sequence, recording the assumptions rather than relying on labels such as “VaR” or “sensitivity.”
Confirm that both figures are loss measures, use the same currency and valuation basis, and refer to the same portfolio or surplus measure.
Match confidence level and time horizon. If time scaling was used, verify the assumptions behind it rather than treating it as an automatic conversion.
List the market factors in each calculation. Determine whether an omitted factor is immaterial or could explain a meaningful part of the gap.
Check whether a first-order sensitivity is adequate for the size of the market moves being measured. Nonlinear positions may need full revaluation or a higher-order approximation.
Compare factor distributions, dependence, and tail treatment. A similar variance does not guarantee a similar VaR.
Common Mistakes and Why the Distinction Matters
A frequent comparison error is to treat a sensitivity number as though it were an independent estimate of the same VaR without checking its construction. A sensitivity calculation may be a useful approximation, but its interpretation depends on its assumptions. A historical or simulation-based VaR may incorporate nonlinear revaluation or a different empirical distribution, so a gap can reflect model design rather than a mistake.
- Comparing different quantiles. A 99% estimate should not be described as agreeing or disagreeing with a 95% estimate until both are placed at a common confidence level.
- Scaling horizons mechanically. Square-root-of-time scaling follows under restrictive assumptions such as independent, identically distributed normal increments and a linear exposure; it is not universal.
- Ignoring curvature. A first-order sensitivity omits nonlinear valuation terms. The omission may be small for modest moves and material for options or large scenarios.
- Comparing unequal factor sets. A measure that omits a market factor cannot be expected to match one that includes it, even if the included-factor calculations are identical.
- Assuming equal variance means equal tail risk. Distribution shape affects quantiles, particularly in the tail. Moment matching alone does not establish quantile agreement.
The purpose of the diagnosis is not to force two methods to produce the same answer. It is to state precisely what drives their difference, which assumptions are responsible, and whether those assumptions are appropriate for the decision being supported. A clear reconciliation is more informative than an unexplained numerical match.
Check Your Understanding
For each problem, state which assumptions or comparison steps control the result. Amounts are in millions of dollars unless stated otherwise.
- A hypothetical portfolio has zero-mean normal daily losses with standard deviation $0.40 million. Daily losses are independent and identically distributed. Find its 5-day, 97.5% VaR using \(z_{0.975}=1.96\). (Answer: \(1.96(0.40)\sqrt{5}=\$1.753\) million.)
- A linear sensitivity model has zero-mean normal daily loss with standard deviation $2.00 million. Find the one-day 95% and four-day 99% VaRs, assuming independent, identically distributed daily losses. Use \(z_{0.95}=1.6449\) and \(z_{0.99}=2.3263\). (Answer: One-day 95% VaR \(=\$3.290\) million; four-day 99% VaR \(=\$9.305\) million. They are not directly comparable because both horizon and confidence level differ.)
- A hypothetical portfolio has \(\Delta V=3{,}000{,}000x-4{,}000{,}000x^2\), with \(x\) uniform on \([-0.01,0.01]\). Find the exact 95% loss VaR and the first-order estimate. (Answer: The loss decreases over the interval, so use the 5th percentile \(x=-0.009\); exact VaR \(=\$27{,}324.00\), first-order estimate \(=\$27{,}000.00\).)
- Independent zero-mean normal loss components have standard deviations $0.30 million and $0.40 million. Find the 99% VaR with both components, the VaR with only the first component, and the difference. Use \(z_{0.99}=2.3263\). (Answer: Both factors: \(\$1.163\) million; first factor only: \(\$0.698\) million; difference: \(\$0.465\) million.)
- A loss is uniform on \([-0.5,0.5]\) million. Compare its exact 95% VaR with a normal estimate having the same mean and variance. Use \(z_{0.95}=1.6449\). (Answer: Exact uniform VaR \(=\$0.450\) million; moment-matched normal estimate \(=1.6449(0.5/\sqrt{3})=\$0.475\) million, rounded to the nearest $0.001 million.)